Numerical solution of equations

Maths Pure 3 · Numerical solution of equations. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

Many equations, such as e^x = 3x, cannot be solved exactly, so we find a root as a decimal instead.

First we locate the root. Write the equation as f(x) = 0. If f is continuous and f(a) and f(b) have opposite signs, there is a root between a and b. This is the change of sign method. A sketch of two graphs, for example y = e^x and y = 3x, shows how many roots there are: one for each crossing point.

Then we improve the value with an iterative formula x_n+1 = F(x_n), made by rearranging the equation into the form x = F(x). Each answer is put back into the formula. If the values settle down, they converge to a root.

Rules to remember

Change of sign: if f is continuous on the interval and f(a), f(b) have opposite signs, a root lies between a and b
The method fails if the curve has a break (an asymptote) in the interval, and it can miss roots when there is an even number of them in the interval
To show a root is 1.53 correct to 2 decimal places, show a change of sign between 1.525 and 1.535
If x_n+1 = F(x_n) converges to α, then α = F(α); rearrange this to find the equation that α satisfies
Keep at least 4 decimal places in each iterate when the answer is wanted to 2 decimal places, and stop when two iterates agree to the accuracy needed
Different rearrangements of the same equation can converge to different roots, or not converge at all
Use radians for any iteration with sin, cos or tan

Common mistake

Students stop the iteration too early or round each iterate to 2 decimal places before using it again. Use the full calculator value (the ANS key) each time, write down each iterate to 4 decimal places, and round only at the end.

Worked example

The equation x³ − 2x − 5 = 0 has a root between 2 and 3. Use x_n+1 = ∛(2x_n + 5) with x₁ = 2 to find the root correct to 2 decimal places.

Check the interval: f(2) = 8 − 4 − 5 = −1 and f(3) = 27 − 6 − 5 = 16, so the sign changes.
x₂ = ∛(2 × 2 + 5) = ∛9 = 2.0801.
x₃ = ∛(2 × 2.0801 + 5) = 2.0924, x₄ = 2.0942, x₅ = 2.0945.
The iterates all round to 2.09 from x₃ onwards.

Answer: The root is 2.09 (2 decimal places)

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